Let $p$ be an odd prime. Show that $\sum_{\beta \in \mathbb{Z}_p^*} \beta^{-1} = \sum_{\beta \in \mathbb{Z}_p^*} \beta = 0$.

== Attempt ==

Given $p$ is prime, $\mathbb{Z}_p = \mathbb{Z}_p^*$, which means all integers from $1$ to $(p - 1)$ are in $\mathbb{Z}_p^*$. Furthermore, every element has a multiplicative inverse, and is itself a multiplicative inverse of some element. Thus,

$\displaystyle\sum_{\beta \in \mathbb{Z}_p^*} \beta^{-1} = \sum_{\beta \in \mathbb{Z}_p^*} \beta = 1 + \dots + (p-1) = \frac{p(p-1)}{2}$

Given $p$ is an odd prime, $2 \mid (p-1)$. Therefore,

$\displaystyle \lbrack \frac{p(p-1)}{2} \rbrack_p = \lbrack p \rbrack_p = \lbrack 0 \rbrack_p$

$\Box$